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A string of total length $l$ is threaded through a hole in the centre of a frictionless horizontal table and then released. How would one go about finding the vertical equation of motion for this string, using Newtonian mechanics rather than the Lagrangian?

In the textbook I am reading the equation of motion is stated as $\rho yg=\rho l\frac{d^2y}{dt^2}$, but I'm not too sure about that. The term on the right hand side implies that the centre of mass of the string is accelerating vertically, but that's not always true. Also, since tension in the string is not constant in this case (it must vary for the horizontal string to accelerate) it shouldn't cancel out everywhere, so shouldn't tension be somewhere in the equation of motion also?

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  • $\begingroup$ Hint: You should start by writing the equation of motion from $F = m a$. $\endgroup$ – Ertxiem Mar 27 at 11:44
  • $\begingroup$ I got the vertical tension gradient as $\rho(\frac{d^2 y}{dt^2}-g)$. $\endgroup$ – user213933 Mar 27 at 11:56
  • $\begingroup$ Well the equation of motion required is of vertical direction not only of centre of mass of the system which may be in horizontal direction.So the equation given in your textbook is of the part of string in vertical direction. $\endgroup$ – user213933 Mar 27 at 12:07
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The mass of the string hanging below the table is $\rho y$ where $\rho$ is the mass per unit length (we are assuming that the string is uniform here). So the force on the string is $\rho yg$.

There is a normal force from the table but this is equal and opposite to the weight of the string that remains on the table, so we can ignore this.

The acceleration of the string is $\frac{d^2y}{dt^2}$. The mass of the whole string is $\rho l$ so applying Newton's second law to the string as a whole gives

$\rho l \frac{d^2y}{dt^2} = \rho yg$

You don't need to worry about whether the centre of mass of the string is on the table or below the table at any point in time - this equation will apply in either case.

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  • $\begingroup$ The string tension will vary from $\rho gy$ at the edge of the table to zero at both the trailing end and the dangling end.. $\endgroup$ – mike stone Mar 27 at 13:02
  • $\begingroup$ @mikestone Good point. I have edited my answer. $\endgroup$ – gandalf61 Mar 27 at 13:46
  • $\begingroup$ "Newton's second law to the string as a whole gives..." - this is the bit that is concerning me. The string as a whole isn't accelerating downwards if the CoM is level on the table, so how can we apply Newton's second law in this way? $\endgroup$ – Pancake_Senpai Mar 27 at 14:03
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    $\begingroup$ @Pancake_Senpai If you don't want to apply Newton II then use an energy argument: $\frac 1 2 \rho l v^2 = \frac 1 2 \rho y^2 g$ so $\rho l v \frac{dv}{dt} = \rho y v g$. $\endgroup$ – gandalf61 Mar 27 at 14:22
  • $\begingroup$ Thanks. That argument makes a lot of sense. I'm still not sure about the application of N2 though, but the energy argument is much cleaner so I'll go with that. $\endgroup$ – Pancake_Senpai Mar 27 at 17:22

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